Question

In triangle $ ABC $, $ 2A + C = 300^\circ $. If the circumradius is 8 times the inradius, then $ \sin\frac{C}{2} = ? $

• A. \( \frac{1}{2} \)
• B. \( \frac{1}{4} \)
• C. \( \frac{3}{4 + \sqrt{3}} \)
• D. \( \frac{5}{4} \)

Hint:

Use identity: \( \frac{R}{r} = \frac{1}{4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}} \)

Answer 1

The Correct Option is B

Let \( A + B + C = 180^\circ \Rightarrow B = 180^\circ - A - C \) Given: \[ 2A + C = 300^\circ \Rightarrow A = \frac{300^\circ - C}{2} \] We use: \[ \frac{R}{r} = \frac{abc}{4rs} \cdot \frac{1}{r} = \frac{abc}{4r^2s} \] But simpler identity: \[ \frac{R}{r} = \frac{1}{4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}} = 8 \Rightarrow 4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2} = \frac{1}{8} \] Since angles are related, substitution leads to: \[ \sin\frac{C}{2} = \frac{1}{4} \]